Let $K/\mathbb{Q}$ be an imaginary quadratic field, $m\ge 1$ be a positive integer and let $\mathcal{O}=\mathbb{Z}+m\mathcal{O}_K\subset \mathcal{O}_K$ be the unique order of index (equivalently, conductor) $m$. I want to know whether the "quotient cancellation property", by which I mean the existence of an isomorphism (as $\mathcal{O}$-modules)

$$\frac{\mathfrak{a}\mathfrak{b}}{\mathfrak{a}\mathfrak{c}}\simeq \frac{\mathfrak{b}}{\mathfrak{c}},$$ holds whenever $\mathfrak{a}$, $\mathfrak{b}\supset \mathfrak{c}$ are invertible fractional $\mathcal{O}$-ideals.

I am able to show it when the quotient $\frac{\mathfrak{b}}{\mathfrak{c}}$ has order relatively prime to $m$. In this case, one may scale both $\mathfrak{b}$ and $\mathfrak{c}$ by some $\mu\in K^{\times}$ to get integral $\mathcal{O}$-ideals which are both relatively prime to the conductor $m$ of $\mathcal{O}$. Doing the same for $\mathfrak{a}$, we can thus assume $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ to be integral, prime-to-$m$, $\mathcal{O}$-ideals.

Notice that one has a multiplicative bijection between the monoid of integral, prime-to-$m$, $\mathcal{O}_K$-ideals and the monoid of integral, prime-to-$m$, $\mathcal{O}$-ideals (as shown in Cox's Primes of the form $x^2+ny^2$, Proposition 7.20 or K. Conrad's note The conductor ideal, Theorem 3.8) given by intersection with $\mathcal{O}$ on one side and by multiplication with $\mathcal{O}_K$ on the other side. Making use of this, and of the "quotient cancellation property" available in the maximal order $\mathcal{O}_K$ (see this topic Cancellation in quotient of fractional ideals for instance), one can get the job done.

However when $|\frac{\mathfrak{b}}{\mathfrak{c}}|$ is no longer prime to $m$, one cannot multiply $\mathfrak{b}$ and $\mathfrak{c}$ by the same scalar to get prime-to-$m$ integral $\mathcal{O}$-ideals and something else is needed to prove (or disprove) the quotient cancellation property.

A related question is the following: if $N$ is square-free with $(N,m)\neq 1$, can we find examples of such invertible fractional $\mathcal{O}$-ideals $\mathfrak{b}\supset \mathfrak{c}$ such that $$\mathfrak{b}/\mathfrak{c}\simeq \mathbb{Z}/N\mathbb{Z} \,\, ?$$

  • $\begingroup$ Anyone, any idea? $\endgroup$ – Yoël Feb 19 '19 at 8:07
  • $\begingroup$ If $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}$ are invertible, isn't $\mathfrak{a} \mathfrak{b}/\mathfrak{a} \mathfrak{c} = \mathfrak{b} \mathfrak{c}^{-1} \mathfrak{a}\mathfrak{a}^{-1}$ equal to $\mathfrak{b} \mathfrak{c}^{-1}$? $\endgroup$ – Bart Michels Sep 17 '19 at 9:10

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